The coordinates of the point at which the XZ- | vedclass

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(B) Let the $XZ$-plane divide the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ in the ratio $k:1$ at point $P$. Since the point $P$ lies on the

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$(7, 0, 1)$

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(B) Let the $XZ$-plane divide the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ in the ratio $k:1$ at point $P$. Since the point $P$ lies on the $XZ$-plane,its $y$-coordinate must be $0$. The coordinates of $P$ are given by the section formula: $P = \left( \frac{k(1) + 1(-2)}{k+1}, \frac{k(2) + 1(3)}{k+1}, \frac{k(3) + 1(4)}{k+1} \right)$. Setting the $y$-coordinate to $0$: $\frac{2k + 3}{k+1} = 0 \implies 2k + 3 = 0 \implies k = -\frac{3}{2}$. Now,substitute $k = -\frac{3}{2}$ into the $x$ and $z$ coordinates: $x = \frac{-\frac{3}{2}(1) - 2}{-\frac{3}{2} + 1} = \frac{-\frac{7}{2}}{-\frac{1}{2}} = 7$. $z = \frac{-\frac{3}{2}(3) + 4}{-\frac{3}{2} + 1} = \frac{-\frac{9}{2} + 4}{-\frac{1}{2}} = \frac{-\frac{1}{2}}{-\frac{1}{2}} = 1$. Thus,the coordinates of the point are $(7, 0, 1)$.

The coordinates of the point at which the $XZ$-plane divides the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ are:

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